EXPLANATION OF RUBIDIUM IONIZATIONS
By Prof. L. Kaliambos ( Natural Philosopher in New Energy) May 30, 2015 Rubidium is a chemical element with symbol Rb and atomic number 37.However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Rubidium including the following ground state electron configuration: 1s22s22p63s23p63d104s24p65s1 . According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of rubidium (from (E1 to E9) are the following: E1 = 4.17, E2 = 27.285, E3 = 40, E4 = 52.6, E5 = 71 , E6 = 84.4, E7 = 99.2 , E8 = 136 and E9 = 150 . Firstly we see that - E1 = - 4.17 = E(5s1) Here the E(5s1) represents the binding energy of the one outermost electron (5s1). Then we observe that - ( E2 +… + E7 ) = - 374.485 = E(4p6) Whereas - ( E8 + E9 ) = - 286 = E(4s2) It is of interest to note that in the absence of data (from E10 to E37 ) one can write the following theoretical ionizations related to the ground state energies: -( Ε10 +…+ E19 ) = E(3d10) - ( E20 +…+ E25 ) = E(3p6). -(E26 + E27) = E(3s2) -(E28 +…+ E33 ) = E(2p6) -(E34 + E35) = E(2s2) -(E36 + E37) = E(1s2) Such theoretical ionization energies are analogous to the experimental values of the ionizations of copper.( See my EXPLANATION OF COPPER IONIZATIONS ). For understanding better the ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. EXPLANATION OF - E1 = - 4.17 = E(5s1) Here the E(5s1) represents the binding energy of the one outermost electron (5s1) given by applying the Bohr formula.' '''The charges (-36e) of the inner electrons (1s22s22p63s23p63d10 4s2 4p6) screen the nuclear charge (+37e) and for a perfect screening we would have ζ = 1. However the electron of 5s1 penetrates the 4p6 and lead to the deformations of spherical electron clouds. Thus ζ > 1. Under this condition we may write Ε1 = 4.17 eV = -E(5s1) = - (-13.6057)ζ2 / n2 Since n = 5 we get ζ = 2.77 > 1 . '''EXPLANATION OF - ( E2 +… + E7 ) = - 374.485 = E(4p6)' The charges (-30e) of the electrons (1s22s22p63s23p63d104s2 ) screen the nuclear charge (+37e) and for a perfect screening we would have an effective ζ = 7. However the six electrons (4p6 ) repel the electrons of 4s2 leading to the deformation of spherical electron clouds. Thus ζ > 7. Under this condition for calculating the E(4p6) of three pairs with opposite spin we may apply my formula of 2008. Thus we write ( E2 + …+ E7 ) = 374.485 eV = -E(4p6) = - 3+ (16.95)ζ - 4.1 / n2 Since n = 4 the above equation is written as 5.1 ζ2 - 3.18 ζ - 373.7 = 0 Then solving for ζ we get ζ = 8.88 > 7 . EXPLANATION OF - ( Ε8 + E9 ) = - 286 = E(4s2) ''' Here the E(4s2) represents the binding energy of the 2 electrons with opposite spin given by applying my formula of 2008. '''The charges (-28e) of the inner electrons (1s22s22p63s23p63d10 ) screen the nuclear charge (+37e) and for a perfect screening we would have ζ = 9. However the electrons of 4s2 penetrate the 3d10 and lead to the deformations of spherical electron clouds. Thus ζ > 9. Under this condition we may write ( Ε8 + E9 ) = 286 eV = -E(4s2) = - + (16.95)ζ - 4.1 / n2 Since n = 4 the above equation could be written as 1.7ζ2 - 1.06ζ - 285.74 = 0 Then solving for ζ we get ζ = 13.28 > 9 . Note that the two electrons of opposite spin (4s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. Category:Fundamental physics concepts